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我試過的一切都給了我錯誤的輸出值。我甚至複製了C代碼並將它們改爲使它們能夠在python中工作，而且我仍然得到錯誤的輸出。哪裏不對？我的ripemd160 python代碼有什麼問題？
import os, math
def makehex(value,size=8):
value = hex(value)[2:]
if value[-1] == 'L':
value = value[0:-1]
while len(value)<size:
value = '0' + value
return value
def makebin(value,size=32):
value = bin(value)[2:]
while len(value)<size:
value = '0' + value
return value
def ROL(value,n):
return (value << n) | (value >> 32-n)
def little_end(string,base = 16):
t = ''
if base == 2:
s= 8
if base == 16:
s = 2
for x in range(len(string)/s):
t = string[s*x:s*(x+1)] + t
return t
def F(x,y,z,round):
if round<16:
return x^y^z
elif 16<=round<32:
return (x & y) | (~x & z)
elif 32<=round<48:
return (x | ~y)^z
elif 48<=round<64:
return (x & z) | (y & ~z)
elif 64<=round:
return x^(y | ~z)
def RIPEMD160(data):
# constants
h0 = 0x67452301; h1 = 0xEFCDAB89; h2 = 0x98BADCFE;h3 = 0x10325476; h4 = 0xC3D2E1F0
k = [0, 0x5A827999, 0x6ED9EBA1, 0x8F1BBCDC, 0xA953FD4E]
kk = [0x50A28BE6, 0x5C4DD124, 0x6D703EF3, 0x7A6D76E9,0]
s = [ 11,14,15,12,5,8,7,9,11,13,14,15,6,7,9,8,
7,6,8,13,11,9,7,15,7,12,15,9,11,7,13,12,
11,13,6,7,14,9,13,15,14,8,13,6,5,12,7,5,
11,12,14,15,14,15,9,8,9,14,5,6,8,6,5,12,
9,15,5,11,6,8,13,12,5,12,13,14,11,8,5,6]
ss= [ 8,9,9,11,13,15,15,5,7,7,8,11,14,14,12,6,
9,13,15,7,12,8,9,11,7,7,12,7,6,15,13,11,
9,7,15,11,8,6,6,14,12,13,5,14,13,13,7,5,
15,5,8,11,14,14,6,14,6,9,12,9,12,5,15,8,
8,5,12,9,12,5,14,6,8,13,6,5,15,13,11,11]
r= range(16) + [ 7, 4, 13, 1, 10, 6, 15, 3, 12, 0, 9, 5, 2, 14, 11, 8,
3, 10, 14, 4, 9, 15, 8, 1, 2, 7, 0, 6, 13, 11, 5, 12,
1, 9, 11, 10, 0, 8, 12, 4, 13, 3, 7, 15, 14, 5, 6, 2,
4, 0, 5, 9, 7, 12, 2, 10, 14, 1, 3, 8, 11, 6, 15, 13]
rr = [ 5, 14, 7, 0, 9, 2, 11, 4, 13, 6, 15, 8, 1, 10, 3, 12,
6, 11, 3, 7, 0, 13, 5, 10, 14, 15, 8, 12, 4, 9, 1, 2,
15, 5, 1, 3, 7, 14, 6, 9, 11, 8, 12, 2, 10, 0, 4, 13,
8, 6, 4, 1, 3, 11, 15, 0, 5, 12, 2, 13, 9, 7, 10, 14,
12, 15, 10, 4, 1, 5, 8, 7, 6, 2, 13, 14, 0, 3, 9, 11]
# md4 padding + preprocessing
temp = ''
for x in data:
temp += makebin(ord(x),8)
length = len(temp)%2**64
temp +='1'
while len(temp)%512!=448:
temp+='0'
input = temp
temp = ''
for x in range(len(input)/32):
temp += little_end(input[32*x:32*(x+1)],2)
input = temp
temp = makebin(length,64)
input += temp[32:]+temp[:32]
t = len(input)/512
# the rounds
for i in range(t):
# i called the parallel round variables 2x the other round variable: a -> aa
a = aa = h0; b = bb = h1; c = cc = h2; d = dd = h3; e = ee = h4
X = input[512*i:512*(i+1)]
X = [int(X[32*x:32*(x+1)],2) for x in range(16)]
for j in range(80):
T = (a+ ROL((F(b, c, d, j) + X[r[j]] + k[j/16])%2**32,s[j])+e)%2**32
c = ROL(c, 10)
a = e; e = d; d = c; c = b; a = T
T = (aa+ ROL((F(bb,cc,dd,79-j) + X[rr[j]] + kk[j/16])%2**32,ss[j])+ee)%2**32
cc = ROL(cc,10)
aa = ee; ee = dd; dd = cc; cc = bb; aa = T
T = (h1+c+dd)%2**32
h1 = (h2+d+ee)%2**32
h2 = (h3+e+aa)%2**32
h3 = (h4+a+bb)%2**32
h4 = (h0+b+cc)%2**32
h0 = T
return little_end(makehex(h0))+little_end(makehex(h1))+little_end(makehex(h2))+little_end(makehex(h3))+little_end(makehex(h4))
data = RIPEMD160('')
print data,data =='9c1185a5c5e9fc54612808977ee8f548b2258d31' # its always false
TL; DR 但是初步瞭解，我猜ROL應該是r編輯一個32位數字？！如果是這樣，'...＆0xFFFFFFFF'丟失。無論如何，hashlib支持RIPEMD160（'hashlib.new（「ripemd160」）'）。 – AndiDog 2010-01-23 18:25:30
我想象很多問題來自整數算術。在C中，算法將依賴於32位溢出。在Python中，整數不會溢出，它們只會被提升爲long。 – bobince 2010-01-23 18:37:01
我已經修改了值，所以我不需要＆ffffff。 – calccrypto 2010-01-23 19:16:59